sphere???????
Q. A sphere has a hole six inches long drilled right through it. What is the volume remaining
Asked by bo nidle - Sat Mar 17 16:38:03 2007 - - 11 Answers - 2 Comments
A. Methinks this is a trick ! Is this a proper question ? If so, the dimensions of the sphere and cylinder must be totally irrelevant. So, let's suppose that the cylinder is infinitely thin - i.e. zero radius, so the sphere must have a diameter of six inches. Its volume is (4/3) .3 = 36 cubic inches.
Answered by sumzrfun - Sat Mar 17 16:53:48 2007
Q. A sphere has a hole six inches long drilled right through it. What is the volume remaining
Asked by bo nidle - Sat Mar 17 16:38:03 2007 - - 11 Answers - 2 Comments
A. Methinks this is a trick ! Is this a proper question ? If so, the dimensions of the sphere and cylinder must be totally irrelevant. So, let's suppose that the cylinder is infinitely thin - i.e. zero radius, so the sphere must have a diameter of six inches. Its volume is (4/3) .3 = 36 cubic inches.
Answered by sumzrfun - Sat Mar 17 16:53:48 2007
What does the sphere of my prescription say about my visual acuity?
Q. I would like to get an idea of what my visual acuity is. People frequently notice how thick one of my lenses is and ask, but I don't really have an answer. The sphere of my glasses prescription (for one eye) says -5.00. If there's no definite number for my VA, could somebody give my an idea? And these doesn't really matter, but the prescription also says polycarbonate. I think that's a type of lens? What makes it different from anything else? And the sphere for my other eye is only -0.25. What causes such a contrast between eyes? Thanks!
Asked by Katie - Fri Jan 22 14:37:53 2010 - - 1 Answers - 0 Comments
A. "The sphere of my glasses prescription (for one eye) says -5.00. If there's no definite number for my VA, could somebody give my an idea?" worse than 20/400. our office (& most) dont routinely measure any worse than that "And these doesn't really matter, but the prescription also says polycarbonate. I think that's a type of lens?" yes thats a common/popular lens material. "What makes it different from anything else? its cheap & impact resistant. "And the sphere for my other eye is only -0.25. What causes such a contrast between eyes?" whew that is quite a bit of difference. as for what "caused" it... either 1 of your eyes is a little larger than the other (front-to-back...axial length) or one of your corneas is "steeper" than… [cont.]
Answered by princeidoc - Fri Jan 22 15:17:34 2010
Q. I would like to get an idea of what my visual acuity is. People frequently notice how thick one of my lenses is and ask, but I don't really have an answer. The sphere of my glasses prescription (for one eye) says -5.00. If there's no definite number for my VA, could somebody give my an idea? And these doesn't really matter, but the prescription also says polycarbonate. I think that's a type of lens? What makes it different from anything else? And the sphere for my other eye is only -0.25. What causes such a contrast between eyes? Thanks!
Asked by Katie - Fri Jan 22 14:37:53 2010 - - 1 Answers - 0 Comments
A. "The sphere of my glasses prescription (for one eye) says -5.00. If there's no definite number for my VA, could somebody give my an idea?" worse than 20/400. our office (& most) dont routinely measure any worse than that "And these doesn't really matter, but the prescription also says polycarbonate. I think that's a type of lens?" yes thats a common/popular lens material. "What makes it different from anything else? its cheap & impact resistant. "And the sphere for my other eye is only -0.25. What causes such a contrast between eyes?" whew that is quite a bit of difference. as for what "caused" it... either 1 of your eyes is a little larger than the other (front-to-back...axial length) or one of your corneas is "steeper" than… [cont.]
Answered by princeidoc - Fri Jan 22 15:17:34 2010
How to calculate the volume of a sphere based on percentage of surface area?
Q. How would you calculate the volume of a sphere based off of a percentage of the surface area if the volume of the sphere is known but only in terms of the surface area? Example: A sphere is measured to displace 280,000 cm^3 of water but its surface area is only submerged 72% under water. What is the total volume of the sphere? Please show work and explain and you will receive points. Could you please show work on how you obtained your answer so I can see if it is right? Thank you.
Asked by C_Rock136 - Fri Feb 22 16:49:04 2008 - - 2 Answers - 0 Comments
A. First the formula for the volume of a sphere is: V = (4/3)(pi)(r^3), and the area of a sphere is: A = 4*(pi)(r^2), which makes it possible to rewrite the two formulae as: V = 4/3(pi)(r^3) = 1/3(A*r) and A = 3(V/r) Now, i know that having 72% of the sphere submerged is multiplying the A by .72. 28000 = 1/3(.72A*r) Solving for A, you get: A = 3*28000/(r*.72) and then equate that to A = (3V)/r (3*28000)/(.72r) = (3V)/(r) 28000/.72 = V V = 388,888.888 cm^3
Answered by NBL - Fri Feb 22 17:57:44 2008
Q. How would you calculate the volume of a sphere based off of a percentage of the surface area if the volume of the sphere is known but only in terms of the surface area? Example: A sphere is measured to displace 280,000 cm^3 of water but its surface area is only submerged 72% under water. What is the total volume of the sphere? Please show work and explain and you will receive points. Could you please show work on how you obtained your answer so I can see if it is right? Thank you.
Asked by C_Rock136 - Fri Feb 22 16:49:04 2008 - - 2 Answers - 0 Comments
A. First the formula for the volume of a sphere is: V = (4/3)(pi)(r^3), and the area of a sphere is: A = 4*(pi)(r^2), which makes it possible to rewrite the two formulae as: V = 4/3(pi)(r^3) = 1/3(A*r) and A = 3(V/r) Now, i know that having 72% of the sphere submerged is multiplying the A by .72. 28000 = 1/3(.72A*r) Solving for A, you get: A = 3*28000/(r*.72) and then equate that to A = (3V)/r (3*28000)/(.72r) = (3V)/(r) 28000/.72 = V V = 388,888.888 cm^3
Answered by NBL - Fri Feb 22 17:57:44 2008
What is the maximum charge that a metallic sphere filled with charged drops of water can accumulate?
Q. The sphere is hollow and has this tiny hole on the top, where the drops may enter. The sphere's radii is R. The sphere is made of methal (conductor). The drops of water have a volume of V, and its charge is Q. What is the net charge that one can accumulate on the sphere?
Asked by Samuel C - Thu May 1 10:27:35 2008 - - 1 Answers - 0 Comments
A. You can put n drops into the sphere where n = 4*pi*R^3/(3V) That means you have a total charge Qt = nQ =4*pi*R^3*Q/(3*V) The electric field normal to the inside surface of the sphere has to go to zero when it enters the metal. So there has to be an induced charge given by: E = nQ/(4*pi*e0*R^2) = s/e0 where s = surface charge density and e0 = 8.85x10^-12 C^2/*nt-m^2) Clean this up by substituting for n QR/(3*V) = s Now s is charge over the surface of the sphere so the induced charge Qi is Qi = -s*4*pi*R^2 "-" because the opposite charges are drawn to the inside surface. THen Q*R/(3*V) = -Qi/(4*pi*R^2) ---> Qi =-Q 4*pi*R^3/(3*V) = nQ Now this charge is balanced by an opposite charge on the other side of the sphere. So… [cont.]
Answered by nyphdinmd - Thu May 1 10:49:27 2008
Q. The sphere is hollow and has this tiny hole on the top, where the drops may enter. The sphere's radii is R. The sphere is made of methal (conductor). The drops of water have a volume of V, and its charge is Q. What is the net charge that one can accumulate on the sphere?
Asked by Samuel C - Thu May 1 10:27:35 2008 - - 1 Answers - 0 Comments
A. You can put n drops into the sphere where n = 4*pi*R^3/(3V) That means you have a total charge Qt = nQ =4*pi*R^3*Q/(3*V) The electric field normal to the inside surface of the sphere has to go to zero when it enters the metal. So there has to be an induced charge given by: E = nQ/(4*pi*e0*R^2) = s/e0 where s = surface charge density and e0 = 8.85x10^-12 C^2/*nt-m^2) Clean this up by substituting for n QR/(3*V) = s Now s is charge over the surface of the sphere so the induced charge Qi is Qi = -s*4*pi*R^2 "-" because the opposite charges are drawn to the inside surface. THen Q*R/(3*V) = -Qi/(4*pi*R^2) ---> Qi =-Q 4*pi*R^3/(3*V) = nQ Now this charge is balanced by an opposite charge on the other side of the sphere. So… [cont.]
Answered by nyphdinmd - Thu May 1 10:49:27 2008
How fast is the sphere moving when it reaches the bottom of the incline?
Q. A solid sphere of uniform density rolls without slipping from the top to the bottom of a 30 incline, a distance 2.6m. The sphere has a mass of 2.5kg and a radius of 28cm. It starts from rest at the top of the incline.
Asked by hikari takishima - Sat Sep 20 09:11:44 2008 - - 3 Answers - 0 Comments
A. since the sphere rolls without slipping,, therefore v=w*r also height of the incline=h*sin30=1.3m so conserving energy we have Mgh = 1/2Mv^2 + 1/2 I w^2 also moment of inertia for solid sphere is 2/5Mr^2 this will get simplified to : Mgh = 7/10 Mv^2 v^2 = 10/7 gh v= 18.5 m/s
Answered by arpit g - Sat Sep 20 09:37:57 2008
Q. A solid sphere of uniform density rolls without slipping from the top to the bottom of a 30 incline, a distance 2.6m. The sphere has a mass of 2.5kg and a radius of 28cm. It starts from rest at the top of the incline.
Asked by hikari takishima - Sat Sep 20 09:11:44 2008 - - 3 Answers - 0 Comments
A. since the sphere rolls without slipping,, therefore v=w*r also height of the incline=h*sin30=1.3m so conserving energy we have Mgh = 1/2Mv^2 + 1/2 I w^2 also moment of inertia for solid sphere is 2/5Mr^2 this will get simplified to : Mgh = 7/10 Mv^2 v^2 = 10/7 gh v= 18.5 m/s
Answered by arpit g - Sat Sep 20 09:37:57 2008
How to figure out the diameter of a sphere from the surface area of a sphere?
Q. I'm stuck on a few problems :( I need to figure out the diameter of a sphere from the surface area of a sphere? For an example, find the diameter of a sphere if the surface area of the sphere is 64pi inches squared. Thanks!
Asked by Kermit - Sun Sep 13 22:01:39 2009 - - 2 Answers - 0 Comments
A. SA = 4(pi)r^2 64(pi)=4(pi)r^2 16=r^2 4=r d=2r d=8
Answered by butterfly - Sun Sep 13 22:06:30 2009
Q. I'm stuck on a few problems :( I need to figure out the diameter of a sphere from the surface area of a sphere? For an example, find the diameter of a sphere if the surface area of the sphere is 64pi inches squared. Thanks!
Asked by Kermit - Sun Sep 13 22:01:39 2009 - - 2 Answers - 0 Comments
A. SA = 4(pi)r^2 64(pi)=4(pi)r^2 16=r^2 4=r d=2r d=8
Answered by butterfly - Sun Sep 13 22:06:30 2009
A sphere, radius r, contains water to a depth h. What is the formula for volume of water to depth?
Q. The sphere is hollow (obviously). Imagine callibrating a dipstick to show the volume of water in a spherical tank.
Asked by johndt47 - Tue Jan 15 13:06:06 2008 - - 5 Answers - 1 Comments
A. V = r/6(3 + h ) where h = depth and r = radius at the water's surface Given the sphere's radius R r = (h - 2Rh) EDIT: Thank you MVB, I did err, twice. The first formula should be V = h/6(3r + h ) and the last formula should be r = (2Rh - h ) Sorry
Answered by Catenary - Tue Jan 15 13:51:00 2008
Q. The sphere is hollow (obviously). Imagine callibrating a dipstick to show the volume of water in a spherical tank.
Asked by johndt47 - Tue Jan 15 13:06:06 2008 - - 5 Answers - 1 Comments
A. V = r/6(3 + h ) where h = depth and r = radius at the water's surface Given the sphere's radius R r = (h - 2Rh) EDIT: Thank you MVB, I did err, twice. The first formula should be V = h/6(3r + h ) and the last formula should be r = (2Rh - h ) Sorry
Answered by Catenary - Tue Jan 15 13:51:00 2008
How much work is required to completely submerge the sphere?
Q. A sphere has a negligible weight and is flowing in a large freshwater lake. How much work is required to completely submerge the sphere? Proof of calculation needed + explanation
Asked by lime_greenluv - Mon Mar 9 07:42:43 2009 - - 3 Answers - 0 Comments
A. Mass of water displaced x time to displace this water. I'm not doing it for you, as you wouldn't learn anything, would ya?
Answered by Yawn - Mon Mar 9 07:47:51 2009
Q. A sphere has a negligible weight and is flowing in a large freshwater lake. How much work is required to completely submerge the sphere? Proof of calculation needed + explanation
Asked by lime_greenluv - Mon Mar 9 07:42:43 2009 - - 3 Answers - 0 Comments
A. Mass of water displaced x time to displace this water. I'm not doing it for you, as you wouldn't learn anything, would ya?
Answered by Yawn - Mon Mar 9 07:47:51 2009
How to count the diameter in a sphere?
Q. I need to know how to count the diameter of the base if I cut a sphere in two parts when the cut would occur parallel to the sphere's diameter. I know the diameter of the sphere and need to move up a certain number on its surface (NOT on it's axis). So, how do I count the diameter at that cut? Please, at least point me in the general direction. I need this very much. Spherical cap. That's what I'm talking about.
Asked by Martina - Fri Jan 29 15:28:35 2010 - - 1 Answers - 0 Comments
A. Hint: use similar triangles.
Answered by michaelempeigne - Fri Jan 29 15:37:20 2010
Q. I need to know how to count the diameter of the base if I cut a sphere in two parts when the cut would occur parallel to the sphere's diameter. I know the diameter of the sphere and need to move up a certain number on its surface (NOT on it's axis). So, how do I count the diameter at that cut? Please, at least point me in the general direction. I need this very much. Spherical cap. That's what I'm talking about.
Asked by Martina - Fri Jan 29 15:28:35 2010 - - 1 Answers - 0 Comments
A. Hint: use similar triangles.
Answered by michaelempeigne - Fri Jan 29 15:37:20 2010
What is the difference between a colony, a sphere of influence, a protectorate and a territory?
Q. What is the difference between a colony, a sphere of influence, a protectorate and a territory? Like in history and colonial powers and stuff...anyone?
Asked by E Biz - Sat Apr 12 23:26:51 2008 - - 4 Answers - 0 Comments
A. A Territory is an integral possession of the couhtry, with aspects of self-government but with the executive appointed by the central government. In US history, Territories were expected to become States whenevr they were deemed to have sufficient population. A Protectorate is a sovereign country under the protection of another country. Usuallly this entails not only guaranteeing the protectorate's territory against encroachment by other powers, but some degree of supervision of at least the Protectorate's foreign affairs so that it doesn't provoke a crisis. Some coutries have been "Protected" right out of existance, such as Hawaii. Others have undergone a period of "getting on their feet" and gone onto become fully independent countries,… [cont.]
Answered by Sydney Michelle - Sat Apr 12 23:54:35 2008
Q. What is the difference between a colony, a sphere of influence, a protectorate and a territory? Like in history and colonial powers and stuff...anyone?
Asked by E Biz - Sat Apr 12 23:26:51 2008 - - 4 Answers - 0 Comments
A. A Territory is an integral possession of the couhtry, with aspects of self-government but with the executive appointed by the central government. In US history, Territories were expected to become States whenevr they were deemed to have sufficient population. A Protectorate is a sovereign country under the protection of another country. Usuallly this entails not only guaranteeing the protectorate's territory against encroachment by other powers, but some degree of supervision of at least the Protectorate's foreign affairs so that it doesn't provoke a crisis. Some coutries have been "Protected" right out of existance, such as Hawaii. Others have undergone a period of "getting on their feet" and gone onto become fully independent countries,… [cont.]
Answered by Sydney Michelle - Sat Apr 12 23:54:35 2008
How do I find the volume of a sphere drilled into by a cylinder?
Q. I have a homework problem: A cylindrical drill with radius 5 is used to bore a hole through the center of a sphere of radius 7. Find the volume of the ring shaped solid that remains. I know this involves double integrals, but I have no idea how to go about implementing it. Any help would be appreciated, thank you.
Asked by Fench - Sat May 22 08:56:42 2010 - - 2 Answers - 0 Comments
A. The easiest way is to picture this with cylindric coordinates. Then you can integrate in the z direction The limits would be from - (7^2 - 5^2) to + (7^2 - 5^2) (z at the edges of the hole) each slice has an outer radius of (7^2-z^2) an inner radius of 5 a thickness of dz and an area of ( 7^2 - z^2 - 5^2) integrate ( 7^2 - z^2 - 5^2) dz and get (24 z - z^3/3) evaluate between limits above 2 (7^2-5^2) (24 - (7^2-5^2)/3) = 2 *16* (24)= 492.5 round to 490
Answered by Physicsmom - Sun May 23 17:49:39 2010
Q. I have a homework problem: A cylindrical drill with radius 5 is used to bore a hole through the center of a sphere of radius 7. Find the volume of the ring shaped solid that remains. I know this involves double integrals, but I have no idea how to go about implementing it. Any help would be appreciated, thank you.
Asked by Fench - Sat May 22 08:56:42 2010 - - 2 Answers - 0 Comments
A. The easiest way is to picture this with cylindric coordinates. Then you can integrate in the z direction The limits would be from - (7^2 - 5^2) to + (7^2 - 5^2) (z at the edges of the hole) each slice has an outer radius of (7^2-z^2) an inner radius of 5 a thickness of dz and an area of ( 7^2 - z^2 - 5^2) integrate ( 7^2 - z^2 - 5^2) dz and get (24 z - z^3/3) evaluate between limits above 2 (7^2-5^2) (24 - (7^2-5^2)/3) = 2 *16* (24)= 492.5 round to 490
Answered by Physicsmom - Sun May 23 17:49:39 2010
What is the speed of each sphere when they are very far apart?
Q. The four 1.0 g spheres shown in the figure are released simultaneously and allowed to move away from each other. Four charges are placed in the corners of a square, 1cm apart. All 4 charges have a charge of +10nC. What is the speed of each sphere when they are very far apart?
Asked by bennyb - Sun Feb 4 20:39:12 2007 - - 1 Answers - 0 Comments
A. The daunting questing is what does it mean "very far apart"? The force pushing them apart is a vector superposition of force by 3 other charger on the forth Ft=F1+F2+F3 (vector sum) F1=F2=F3 Ftx=F(cos(0) + cos(45) + cos (90)) Ftx=F + Fcos(45) Fty=F(sin(0) + sin(45) + sin (90)) Fty=Fsin(45) + F F=kq^2/r^2 (we are not out of the woods yet sinceF=f(r) F is a function of distance between them!) Let's look a t velocity V V=at F=ma a=F/m so V=(F/m) t ( simple enough ;)) Now let's do some calculus dF=-2kq^2/r^3 so F= -2kq^2 integral ( 1/r^3dr) [from r1=1cm to r2 =what ever you want] F= kq^2( 1/r2^2 - 1/r1^2) If we assume r2 to be very large F=kq^2(1/r1^2) (exercise in ...) Now just do the arithmetic...
Answered by Edward - Mon Feb 5 10:40:11 2007
Q. The four 1.0 g spheres shown in the figure are released simultaneously and allowed to move away from each other. Four charges are placed in the corners of a square, 1cm apart. All 4 charges have a charge of +10nC. What is the speed of each sphere when they are very far apart?
Asked by bennyb - Sun Feb 4 20:39:12 2007 - - 1 Answers - 0 Comments
A. The daunting questing is what does it mean "very far apart"? The force pushing them apart is a vector superposition of force by 3 other charger on the forth Ft=F1+F2+F3 (vector sum) F1=F2=F3 Ftx=F(cos(0) + cos(45) + cos (90)) Ftx=F + Fcos(45) Fty=F(sin(0) + sin(45) + sin (90)) Fty=Fsin(45) + F F=kq^2/r^2 (we are not out of the woods yet sinceF=f(r) F is a function of distance between them!) Let's look a t velocity V V=at F=ma a=F/m so V=(F/m) t ( simple enough ;)) Now let's do some calculus dF=-2kq^2/r^3 so F= -2kq^2 integral ( 1/r^3dr) [from r1=1cm to r2 =what ever you want] F= kq^2( 1/r2^2 - 1/r1^2) If we assume r2 to be very large F=kq^2(1/r1^2) (exercise in ...) Now just do the arithmetic...
Answered by Edward - Mon Feb 5 10:40:11 2007
What is the sphere's angular velocity at the bottom of the incline?
Q. An 8.40-cm-diameter, 300 sphere is released from rest at the top of a 2.00-m-long, 17.0 incline. It rolls, without slipping, to the bottom. What fraction of its kinetic energy is rotational?
Asked by juan - Wed Apr 1 20:22:20 2009 - - 1 Answers - 0 Comments
A. Here to calculate the angular velocity, and the rotational kinetic energy, you must use: radius = 8.4 cm, not half of that The reason being that the center of rotation for a rolling sphere is the contact point with the incline, NOT the center of the sphere.
Answered by PhysicsDude - Sun Apr 5 18:00:07 2009
Q. An 8.40-cm-diameter, 300 sphere is released from rest at the top of a 2.00-m-long, 17.0 incline. It rolls, without slipping, to the bottom. What fraction of its kinetic energy is rotational?
Asked by juan - Wed Apr 1 20:22:20 2009 - - 1 Answers - 0 Comments
A. Here to calculate the angular velocity, and the rotational kinetic energy, you must use: radius = 8.4 cm, not half of that The reason being that the center of rotation for a rolling sphere is the contact point with the incline, NOT the center of the sphere.
Answered by PhysicsDude - Sun Apr 5 18:00:07 2009
How do you work out the radius of a sphere from the total surface area?
Q. I have maths test coming up and in my revision test I've come across this. We haven't covered it in school and I'm not sure what to do. It wants me to work out the radius of a sphere from the total surface area. The total surface area is 100cm^2. What is the radius? Also, rather than just giving me the answer, can it be worked out and explained step-by-step? Because in the real test different numbers will be used.
Asked by Stephen - Wed Aug 26 04:07:23 2009 - - 1 Answers - 0 Comments
A. Formula for surface area of a sphere pi=3.14 surface area= 4 times pi times r^2 From this formula you can get the value of R which is the radius R= square root of surface area / 4 times pi =Square root of 100 / 4 TIMES 3.14 =Square root of100 / 12.56 =Square root of 7.96 =2.82 cm. therefore, The radius measures 2.82 cm.
Answered by MixterZ - Wed Aug 26 04:22:14 2009
Q. I have maths test coming up and in my revision test I've come across this. We haven't covered it in school and I'm not sure what to do. It wants me to work out the radius of a sphere from the total surface area. The total surface area is 100cm^2. What is the radius? Also, rather than just giving me the answer, can it be worked out and explained step-by-step? Because in the real test different numbers will be used.
Asked by Stephen - Wed Aug 26 04:07:23 2009 - - 1 Answers - 0 Comments
A. Formula for surface area of a sphere pi=3.14 surface area= 4 times pi times r^2 From this formula you can get the value of R which is the radius R= square root of surface area / 4 times pi =Square root of 100 / 4 TIMES 3.14 =Square root of100 / 12.56 =Square root of 7.96 =2.82 cm. therefore, The radius measures 2.82 cm.
Answered by MixterZ - Wed Aug 26 04:22:14 2009
What is a Celestial Sphere and what is this question asking?
Q. Imagine that your teacher gives you a bunch of stars and a black umbrella. With this equipment, you must represent the celestial sphere. On what part of the umbrella would you place the stars? Explain your reasoning. [Please tell me ! :S]
Asked by MelThrills - Tue Jun 2 19:30:12 2009 - - 2 Answers - 0 Comments
A. it sounds as if your teacher is asking you to describe the night sky as a celestial sphere. If thats the case than the stars would be spread all around the inside of the umbrella as they are in the night sky
Answered by gtarczar - Tue Jun 2 19:37:20 2009
Q. Imagine that your teacher gives you a bunch of stars and a black umbrella. With this equipment, you must represent the celestial sphere. On what part of the umbrella would you place the stars? Explain your reasoning. [Please tell me ! :S]
Asked by MelThrills - Tue Jun 2 19:30:12 2009 - - 2 Answers - 0 Comments
A. it sounds as if your teacher is asking you to describe the night sky as a celestial sphere. If thats the case than the stars would be spread all around the inside of the umbrella as they are in the night sky
Answered by gtarczar - Tue Jun 2 19:37:20 2009
How does an energy sphere work?
Q. Why does an energy sphere require two people to work? Elementary science textbook shows 2 students with an energy sphere that lights up when 2 students have eye contact & one is holding it. Elementary science textbook shows 2 students with an energy sphere that lights up when 2 students have eye contact & one is holding it.
Asked by MJT - Tue Aug 5 08:56:06 2008 - - 1 Answers - 0 Comments
A. What is an energy sphere? I suspect it is one of those YouTube fakes. A search of the internet turned up many different references, which one are you talking about? . .
Answered by billrussell42 - Tue Aug 5 09:54:38 2008
Q. Why does an energy sphere require two people to work? Elementary science textbook shows 2 students with an energy sphere that lights up when 2 students have eye contact & one is holding it. Elementary science textbook shows 2 students with an energy sphere that lights up when 2 students have eye contact & one is holding it.
Asked by MJT - Tue Aug 5 08:56:06 2008 - - 1 Answers - 0 Comments
A. What is an energy sphere? I suspect it is one of those YouTube fakes. A search of the internet turned up many different references, which one are you talking about? . .
Answered by billrussell42 - Tue Aug 5 09:54:38 2008
Point on the celestial sphere where the sun is farthest south of the celestial equator?
Q. I'm having a really hard time understand the celestial sphere in reference to the sun. At what point on the celestial sphere will the sun be when it is farthest south of the celestial equator? Also, at what place on the celestial sphere will the sun cross the celestial equator while appearing to move northward?
Asked by Em[ILY] - Fri Apr 16 22:20:06 2010 - - 1 Answers - 0 Comments
A. By definition: 1) December solstice (also called "winter solstice" by people living in the northern hemisphere) 2) March equinox (a.k.a., Vernal equinox -- also "Spring equinox" for people in the northern hemisphere"
Answered by Raymond - Fri Apr 16 22:25:15 2010
Q. I'm having a really hard time understand the celestial sphere in reference to the sun. At what point on the celestial sphere will the sun be when it is farthest south of the celestial equator? Also, at what place on the celestial sphere will the sun cross the celestial equator while appearing to move northward?
Asked by Em[ILY] - Fri Apr 16 22:20:06 2010 - - 1 Answers - 0 Comments
A. By definition: 1) December solstice (also called "winter solstice" by people living in the northern hemisphere) 2) March equinox (a.k.a., Vernal equinox -- also "Spring equinox" for people in the northern hemisphere"
Answered by Raymond - Fri Apr 16 22:25:15 2010
What if a deep space probe slammed into the celestial sphere?
Q. What if Aristotle and Ptolemy were right? What if the Voyagers 1 and 2 crash into the celestial sphere? How would that change our space program and our view of our place in the universe?
Asked by Link - Sat Aug 25 14:53:12 2007 - - 5 Answers - 0 Comments
A. Aristotle and Ptolemy didn't say there was a single celestial sphere. They said there were many spheres (one for each visible solar system object), at varying distances from the Earth, and all concentric. All the way out to the one holding the "stationary" stars. If they were right, then Voyager 2 (the first of the two to launch) would have proved it 30 years ago, when it would have shattered the one containing the Moon. Interestingly enough, that didn't happen. Nor did it happen later when they passed through the "spheres" of Mercury, Venus, the Sun, Mars, Jupiter, etc., etc. Nor when numerous other probes passed through various "spheres" over the twenty years before them. To hold that the solar system is composed of mostly empty… [cont.]
Answered by skeptik - Mon Aug 27 21:18:37 2007
Q. What if Aristotle and Ptolemy were right? What if the Voyagers 1 and 2 crash into the celestial sphere? How would that change our space program and our view of our place in the universe?
Asked by Link - Sat Aug 25 14:53:12 2007 - - 5 Answers - 0 Comments
A. Aristotle and Ptolemy didn't say there was a single celestial sphere. They said there were many spheres (one for each visible solar system object), at varying distances from the Earth, and all concentric. All the way out to the one holding the "stationary" stars. If they were right, then Voyager 2 (the first of the two to launch) would have proved it 30 years ago, when it would have shattered the one containing the Moon. Interestingly enough, that didn't happen. Nor did it happen later when they passed through the "spheres" of Mercury, Venus, the Sun, Mars, Jupiter, etc., etc. Nor when numerous other probes passed through various "spheres" over the twenty years before them. To hold that the solar system is composed of mostly empty… [cont.]
Answered by skeptik - Mon Aug 27 21:18:37 2007
What is the magnitude of the electric field at a point away from the sphere's center?
Q. A nonconducting solid sphere of radius 2.8 cm carries a uniformly distributed positive charge of 6.5 x10-9 Coulombs. Calculate the magnitude of the electric field at a point 1.3 cm away from the center of the sphere.
Asked by Kyle - Wed Feb 3 20:07:12 2010 - - 1 Answers - 0 Comments
A. k Q/V(total) * V(under) /r^2 (where k = 1/4 =9 10^9Nm^2/C^2 ) k Q/( R^2) * r^2/r^2 =k Q/R^2 = 9 10^9 Nm^2/C^2 6.5 x10^-9 C / (2.8*10^(-2)m)^2 = 8.29 x10^-6
Answered by Physicsmom - Fri Feb 5 15:10:19 2010
Q. A nonconducting solid sphere of radius 2.8 cm carries a uniformly distributed positive charge of 6.5 x10-9 Coulombs. Calculate the magnitude of the electric field at a point 1.3 cm away from the center of the sphere.
Asked by Kyle - Wed Feb 3 20:07:12 2010 - - 1 Answers - 0 Comments
A. k Q/V(total) * V(under) /r^2 (where k = 1/4 =9 10^9Nm^2/C^2 ) k Q/( R^2) * r^2/r^2 =k Q/R^2 = 9 10^9 Nm^2/C^2 6.5 x10^-9 C / (2.8*10^(-2)m)^2 = 8.29 x10^-6
Answered by Physicsmom - Fri Feb 5 15:10:19 2010
How was sphere eversion proved possible using rigorous language?
Q. I know that someone proved it was possible. But how? What kind of language and branch of mathematics would you have to use to describe the surface of a manifold? And how was this language used to form a rigorous proof of sphere eversibility (word I made up to describe the ability to turn a sphere inside-out haha)?
Asked by shugo - Sun Jan 4 08:33:35 2009 - - 1 Answers - 0 Comments
A. Topology I read a good proof in Scientific American about 30 years ago. Go to the Scientific American site and buy the reprint it was published in the 70's. sorry i can't remember the issue better, Go to a Good library, Your question is a little deep for this medium
Answered by mathman241 - Sun Jan 4 08:55:52 2009
Q. I know that someone proved it was possible. But how? What kind of language and branch of mathematics would you have to use to describe the surface of a manifold? And how was this language used to form a rigorous proof of sphere eversibility (word I made up to describe the ability to turn a sphere inside-out haha)?
Asked by shugo - Sun Jan 4 08:33:35 2009 - - 1 Answers - 0 Comments
A. Topology I read a good proof in Scientific American about 30 years ago. Go to the Scientific American site and buy the reprint it was published in the 70's. sorry i can't remember the issue better, Go to a Good library, Your question is a little deep for this medium
Answered by mathman241 - Sun Jan 4 08:55:52 2009
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